Note on Distance Magic Products \(G\circ C_4\)

Marcin Anholcer 0 Sylwia Cichacz 0 0 S. Cichacz Faculty of Applied Mathematics, AGH University of Science and Technology , Al . Mickiewicza 30, 30-059 Krakw, Poland A distance magic labeling of a graph G = (V , E ) of order n is a bijection l : V {1, 2, . . . , n} with the property that there is a positive integer k (called magic constant) such that w(x ) = k for every x V . If a graph G admits a distance magic labeling, then we say that G is a distance magic graph. In the case of non-regular graph G, the problem of determining whether there is a distance magic labeling of the lexicographic product G C4 was posted in Arumugam et al. (J Indonesian Math Soc 11-26, 2011). We give necessary and sufficient conditions for the graphs Km,n C4 to be distance magic. We also show that the product C3(t) C4 of the Dutch Windmill Graph and the cycle C4 is not distance magic for any t > 1. All graphs considered in this paper are simple finite graphs. Given a graph G, we denote its order by |G| = n, its vertex set by V (G) and the edge set by E (G). The - neighborhood N (x ) of a vertex x is the set of vertices adjacent to x , and the degree d(x ) of x is |N (x )|, the size of the neighborhood of x . Let w(x ) = yNG (x) l(y) for every x V (G). A distance magic labeling (also called sigma labeling) of a graph G = (V , E ) of order n is a bijection l : V {1, 2, . . . , n} with the property that there is a positive integer k (called magic constant) such that w(x ) = k for every x V . If a graph G admits a distance magic labeling, then we say that G is a distance magic graph (see [13]). The concept of distance magic labeling has been motivated by the construction of magic squares. Finding a distance magic labeling of an r -regular graph turns out to be equivalent to finding equalized incomplete tournament EIT(n, r ) [4]. In an equalized incomplete tournament EIT(n, r ) of n teams with r rounds, each team plays with exactly r other teams and the total strength of the opponents that team i plays is k. Thus, it is easy to observe that finding an EIT(n, r ) is the same as finding a distance magic labeling of any r -regular graph on n vertices. For a survey, we refer the reader to [2]. The following observations were independently proved: Observation 1.1 ([810,13]) Let G be an r -regular distance magic graph on n vertices. Then k = r(n2+1) . The problem of distance magic labeling of r -regular graphs was studied recently (see [14,9,11]). It is interesting that if you blow up an r -regular G graph into some specific p-regular graph, then the obtained graph H is distance magic. More formally, we have the following definition. Definition 1.3 ([7], p. 185) The lexicographic product G H of two graphs G and H is defined on V (G H ) = V (G) V (H ), two vertices (u, x ), (v, y) of G H being adjacent whenever uv E (G), or u = v and x y E (H ). G H is also called the composition of graphs G and H and denoted by G[H ] (see [6]). Miller at al. [9] proved the following results. Theorem 1.5 ([9]) Let r 1, n 3, G be an r -regular graph and Cn the cycle of length n. Then G Cn admits a distance magic labeling if and only if n = 4. Theorem 1.6 ([9]) Let G be an arbitrary regular graph. Then G K n is distance magic for any even n. Shafiq et al. [12] considered distance magic labeling for disconnected graphs and obtained the following theorems. Theorem 1.7 ([12]) Let m 1, n 2 and p 3. Then mC p Kn has a distance magic labeling if and only if either n is even or mnp is odd or n is odd and p 0(mod 4). The following problem was posted in [2]. Proposition 1.8 ([2]) If G is non-regular graph, determine if there is a distance magic labeling of G C4. The Dutch Windmill Graph C (t), also called a friendship graph, is the graph obtained 3 by taking t > 1 copies of the cycle graph C3 with a vertex in common [5]. We show that the product C3(t) C4 is not distance magic for any t > 1. The paper is organized as follows. In the next section we focus on the products of complete bipartite graphs and cycle C4. In the third section we prove that the product of the Dutch Windmill Graph and the cycle C4 cannot be distance magic. 2 The Product Km,n C4 Let Km,n have the vertex partite sets A ={x0, x1, . . . , xn1} and B= {y0, y1, . . . , ym1}. Let C4 = v0v1v2v3v0 and H = Km,n C4. For 0 i n 1 and j = 0, 1, 2, 3, let xij 0, 1, 2, 3, xi A} and B[C4] = {yij : i = 0, 1, . . . , m 1, j = 0, 1, 2, 3, yi B}. The following Lemma holds true. Lemma 2.1 If H = Km,n C4, where 1 m < n is a distance magic graph and k is the magic constant, then the following conditions hold: (1) l(xi0)+l(xi2) = l(xi1)+l(xi3) = a for some constant a for all 0 i n 1 and l(yi0) + l(yi2) = l(yi1) + l(yi3) = b for some constant b for all 0 i m 1, (2) b + 2an = a + 2mb = k and a < b, (3) bm + an = (m + n)(4m + 4n + 1), Proof (1) Notice that: w(xij ) = l(xij+1) + l(xij+3) + im=1 4j=1 l(yij ) for all 0 i n 1, j = 0, 1, 2, 3, where the addition in the superscripts is performed modulo 4. Since the graph H is distance magic we obtain that l(xi0) + l(xi2) = l(xi1) + l(xi3) = a for some constant a for all 0 i n 1. Similarly l(yi0) + l(yi2) = l(yi1) + l(yi3) = b for some constant b for all 0 i m 1. (2) Fact (1) implies that w(xij ) = a + 2bm for all 0 i n 1, j = 0, 1, 2, 3 and w(yij ) = b + 2an for all 0 i m 1, j = 0, 1, 2, 3. As m < n, this implies that a < b. (3) The labeling l is a bijection, so the sum of all labels has to be equal to 2an + 2bm = The following theorem completely characterizes the pairs (m, n), for which Km,n C4 is distance magic. Theorem 2.2 Let m and n be integers such that 1 m < n. Then Km,n C4 is distance magic if and only if the following conditions hold. (1) The numbers are integers. (2) There exist integers p, q, t 1, such that a = b = 4m = qt. Proof First, let us assume that for given m and n, 1 m < n there exist a, b, p, q and t with desired properties. Then the following labeling is distance-magic: If t = 4s for some integer s, then let k(2 p + 2q) + i + 1 for j = 0 l(xkjp+i ) = k(2 p + 2q) + p + q + i + 1 for j = 1, a f (xkjp+2i ) for j = 2, 3, for 0 k t /4 1, i = 0, 1, . . . , p 1, k(2 p + 2q) + p + i + 1 for j = 0, l(ykjq+i ) = k(2 p + 2q) + 2 p + q + i + 1 for j = 1, b f (ykjq+2i ) for j = 2, 3. for 0 k t /4 1, i = 0, 1, . . . , q 1. Observe that the sets of labels of vertices xij and yij for j = 0, 1 do not intersect and their elements are consecutive numbers from the set {1, . . . , 2(m + n)}. And as b a = p + q, also the sets of labels for j = 2, 3 do not intersect and they are consecutive numbers from the set {a (2m + 2n) + q, . . . , a + q 1 = b p 1}. In order to prove that l is a bijection it is enough to show that a + q 1 = 4m + 4n. It is true, as we have: a(n + m) + (b a)m = an + bm = (m + n)(4m + 4n + 1). But on the other hand, (b a)m = ( p + q)m = = (m + n)q, a(n + m) + (m + n)q = (m + n)(4m + 4n + 1) and a + q = 4m + 4n (...truncated)